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The kernel of a linear map $T:V\to W$ is the vector subspace of inputs sent to zero. Linearity makes it a vector subspace: if $T(v)=T(w)=0$, then $T(av+bw)=0$ for all scalars $a,b$. A vector in the kernel represents a change that the map cannot detect, since $T(x+v)=T(x)$ for every input $x$.

For $T:{\mathbb{R}}^3\to {\mathbb{R}}^2$ given by $T(x,y,z)=(x+y,z)$, the zero-output equations are $x+y=0$ and $z=0$. ThusIncreasing the first input while decreasing the second by the same amount leaves the output unchanged. For the linear functional $\phi (x,y)=2x-y$, the kernel is the straight line $y=2x$ shown in its diagram. A linear map is injective exactly when its kernel is the zero vector subspace, because $T(v)=T(w)$ is equivalent to $T(v-w)=0$.

The kernel describes invisible changes, rather than all inputs that produce a specified nonzero output. If $T(x_0)=b$, the corresponding fiber of a function is the affine solution space of a linear equation $x_0+\ker T$. The difference of two inputs in that fiber belongs to the kernel; an input in that fiber need not. In particular, for $b\ne 0$ the fiber cannot be a vector subspace, since it does not contain the zero vector.^[ref]

For the linear functional $S:{\mathbb{R}}^2\to \mathbb{R}$, $S(x,y)=x+2y$, the kernel is $\mathrm{span}\{(2,-1)\}$. Each fiber $S^{-1}(\{c\})$ is the parallel straight line $(c,0)+\mathrm{span}\{(2,-1)\}$. The input vector space is two-dimensional, but the output is one scalar: a whole straight line of inputs collapses to one point of the output real line.